发布网友 发布时间:2024-10-23 19:18
共2个回答
热心网友 时间:2024-10-30 03:57
(1)f(x)=2sinx(sinx+cosx)-1=2sin^2(x)+sin2x=(1-cos2x)+sin2x-1=sin2x-cos2x=√2sin(2x-π/4)
0≤x≤π/2
0≤2x≤π
-π/4≤2x-π/4≤3π/4
-√2/2≤sin(2x-π/4)≤1
-1≤√2sin(2x-π/4)≤-√2
-1≤f(x)≤√2
(2)将y=sinx向右移π/4得到y=sin(x-π/4)再将纵坐标不变,横坐标缩短为原来的一半得到y=sin(2x-π/4)再保持横坐标不变,纵坐标拉伸到原来的)√2倍得到y=√2sin(2x-π/4)
热心网友 时间:2024-10-30 03:59
(1)
f(x)=2sinx(sinx+cosx)-1
=2sin^2(x)+sin2x
=(1-cos2x)+sin2x-1
=sin2x-cos2x
=√2sin(2x-π/4)
-π/4≤(2x-π/4)≤3π/4
-√2/2≤sin(2x-π/4)≤1
-1≤f(x)≤√2
(2)
sin(x)--->右移π/4个单位-->sin(x-π/4)-->横坐标缩短为原来的二分之一倍->sin(2x-π/4)-->
-->纵坐标伸长为原来的根号2倍-->√2sin(2x-π/4)